Atomic Systems (PART I)

Transcription

1 Atomic Systems (PART I) Lecturer: Location: Recommended Text: Dr. D.J. Miller Room 535, Kelvin Building Joseph Black C407 (except 15/1/10 which is in Kelvin 312) Physics of Atoms and Molecules Bransden & Joachain (Prentice Hall, 2 nd Ed.) 1

2 Outline 1. Quantum Mechanics Revision 2. The Schrödinger Equation in a Central Potential 3. The Hydrogen Atom from Schrödinger s Equation 4. Observations of Hydrogen Spectra 5. Fine Splitting of Hydrogen Energy Levels 6. Hydrogen in an External Magnetic Field 2

3 1.1 The Wavefunction 1. Quantum Mechanics Revision A one-particle quantum mechanical system can be described by a classical field known as the wavefunction. wavefunction time position vector The probability density for finding a particle at a position at time is i.e. The probability of finding the particle in a volume V is Conservation of probability 3

4 1.2 The Schrödinger Equation The form of is governed by the Schrödinger Equation and boundary conditions Schrödinger Equation: This is really just a statement of energy conservation. Total energy operator, the Hamiltonian Potential Energy Kinetic energy is the momentum operator Recall 4

5 1.3 Time dependence of the Wavefunction We can use the method of separation of variables to simply things a little. (See Mathematical Methods II if you have chosen to do it.) Lets assume we can write Independent of Independent of Our equation becomes Divide by no time derivative 5

6 no dependence no dependence no or dependence Let s call the constant E (!), then we have 2 equations: 1. another constant 2. These equations are eigenvalue equations they tell us that is an eigenfunction of the Hamiltonian with eigenvalue E. E is the energy of the state. 6

7 Notice that is constant, so above never changes with time. How then can we have a system which changes with time? If the Hamiltonian has multiple eigenvalues (and eigenfunctions) as one would expect, then there exist multiple solutions with different E. A general solution takes the form where is a solution of Although the modulus squared of any particular term is time independent, the interference terms are not: e.g. If then 7

8 1.4 The Time Independent Schrödinger equation Writing out explicitly in Eq.(2) gives the time independent Schrödinger equation: I am going to drop the subscript r from now on and just write as The free Schordinger equation, with V=0 is which has solutions where When V 0 we need to work a little harder! 8

9 1.5 Hermitian operators and orthonormality All the operators which correspond to physical observables, such as energy, momentum and position are Hermitean. An operator A is Hermitian if often this is expressed as Hermitian operators always have real eigenvalues. Imagine is an eigenstate of with eigenvalue A, so that Now but Since these are equal, must be real. 9

10 The eigenstates of Hermitian operators are orthogonal. Imagine are eigenstates of with eigenvalues, i.e. Now but Since these are equal, we must have so either or Putting this together with the probability of finding a particle somewhere being 1 tells us that the wavefunctions are orthonormal: 10

11 2. The Schrodinger Equation in a Central Potential Ultimately in this course, we want to calculate the wavefunction and energy levels of a single electron in a hydrogen-like atom. For an atomic nucleus of charge Ze, the electron sits in a (Coulomb) potential This is an example of a central potential, where the potential depends only on the distance from the source and not the direction. It is useful to consider central potentials more generally first 11

12 2.1 Spherical Polar Coordinates Since we are considering a central potential, it is most efficient to work in spherical polar coordinates In spherical polars, the Laplacian operator is written: This is derived in the course Mathematical Methods II. 12

13 2.2 Separating angular and radial equations The (time-independent) Schrödinger equation becomes We can separate the radial dependence from the angular dependence using the method of separation of variables again. Let s write and substitute it into the equation. 13

14 Divide by and rearrange. no or dependence no r dependence Notice that each side is dependent on different variables, so both must be equal to a constant! We choose to call this constant - we will see why later. 14

15 Radial Equation: We can t solve the radial equation without knowing (we will use it later) Angular Equation: But we can solve the angular part since we have explicitly required the potential to have no angular dependence. 15

16 2.2 Separating the angular equations (Same as on previous slide) Now write, rearrange, divide by and multiply by no dependence no θ dependence So, just like for the previous separation, each side must be constant, say m 2. and 16

17 2.3 Solving the equation for Since the coefficient is a constant, the solution must have a form The general solution is or (usually) more conveniently Notice that m must be an integer since. where A and B are constant (this is where the quantum of quantum mechanics comes from!) 17

18 2.4 Solving the equation for This one is a little bit too hard for now. The solution is an Associated Legendre Polynomial. We will write the general solution as where A is a constant. The l and m here are labels corresponding to the differential equation. We saw earlier that m must be an integer. The associated Legendre function similarly insist that lx 0, and m l. 18

19 Let s first consider the special case of For integer values of, the first few associated Legendre polynomials are where I have written Exercise: Show that 19

20 How do we get these polynomials? Let s write constants Exercise: Plug this into to show that This tells us that For any integer value of one of these series diverges at and one truncates. 20

21 Now we need to apply boundary conditions. In our physical problem,, so varies between -1 and 1. must be finite (and well behaved) in this region. Both series are fine for, but look what happens at. For any non-integer value of both diverge, and the solution doesn t work. For any integer value of one of these series diverges and one truncates. 21

22 For example: For To have a sensible solution at, set (and choose ) For, To have a sensible solution at, set (and choose ) is quantized by boundary conditions, just like. 22

23 More generally we can write the polynomials via (Rodrigues formula): This can be proved by induction, but the proof is a little too long for this course. If you do Mathematical Methods II, you will see it there. For, polynomials can be obtained by differentiation Again, the proof of this is a bit too involved for now, but notice that the highest power of in is. If we differentiate too often then we get zero. 23

24 The solutions to are therefore with These functions are known as spherical harmonics (N is a normalization constant) They are present in the solution of Schrödinger s equation for any central potential. 24

25 The constant N is fixed by requiring an orthonormality condition: measure The infinitesimal volume in spherical polars is so this integration is over just the angular part. This will make things easier later when we need to make sure This condition leads to Exercise: Show the above for 25

26 What happened to our symmetry? The original equation (the time dependent Schrodinger Equation with a central potential) was spherically symmetric, but this solution is not. What happened? The solution we found is not a general solution. Any function of the form will satisfy the equation. we will see the quantum number n soon! We could construct a symmetric solution with appropriate choices of Our choice of the constants and explicitly breaks the symmetry. In reality, we would measure and and the (angular part of the) wavefunction would collapse into one particular choice of It is this collapse of the wavefunction after measurement which is particularly odd about QM 26

27 2.5 The link to Angular Momentum What is the physical significance of and? Classically, angular momentum is defined by To convert to quantum mechanics, all we have to do it turn the observables into operators (first quantisation). QM We know the momentum operator is, and the position operator is (in position space), so angular momentum is In spherical coordinates, this is where we have used basis vectors in spherical polars 27

28 Hopefully, you know by now that the important operators in quantum mechanical discussions of angular momentum are and. What are these in spherical polars? (After some algebra remember to differentiate the basis vectors too!) and Exercise: Prove the above results for and Hint you need to know 28

29 Let s apply these operators to the spherical harmonics: But this is just part of our orginal equation we solved for Therefore and is an eigenstate of with eigenvalue 29

30 Similarly So is also an eigenstate of the operator with eigenvalue Therefore, by definition is the angular part of a state with angular momentum of magnitude and z-component. Exercise: Can you derive the commutators of, and? 30

31 3. The Hydrogen Atom from Schrödinger s Equation 3.1 The Reduced Mass and the Centrifugal Barrier The Hamiltonian for a single electron orbiting a nucleus of charge Ze is with However, this assumes that the nucleus is at rest. In actuality the electron and nucleus will orbit around their joint centre-of-mass and the total energy is better described by: 31

32 Let s rewrite this in terms of the position vector of the centre-of-mass and the vector between the nucleus and the electron This gives: this term is the energy of motion of centre-of-mass total mass reduced mass this term is the energy of relative motion Exercise: Derive this Hamiltonian. 32

33 However, if there is no external force on the atom, the centre-of-mass will move with a constant velocity and the first term (the energy associated with this motion) is a constant. Since we can always ignore a constant (e.g. the value of V(0) is arbitrary), we can ignore this term and have a Schrödinger equation: This is exactly like before, except The angular equation had no dependence on the mass, so everything we did before carries over. All we need to do now is solve the radial equation: with 33

34 3.2 The Centrifugal Barrier We can make this equation look more familiar by defining a new function So, And the radial equation, becomes 34

35 Notice that this is a one-dimensional Schrödinger equation with a modified potential this extra term is known as the centrifugal barrier It is biggest for small r, so has the effect of pushing the wavefunction outwards (exactly like you would expect) Notice that states with E>0 will be free, and only states with E<0 are bound inside the atom. 35

36 3.3 Solving the Radial Equation Our radial equation is To make the maths a wee bit cleaner, let s make the following redefinitions: Then,. Finally we write and to obtain 36

37 This equation has solutions that contain associated Laguerre polynomials Exercise: Plug the above result into the radial equation to recover the associated Laguerre equation, with and Let s take the simpler case of (the Laguerre Equation) and solve it using a power series (just like for the Legendre polynomials) Beware! My definition of Laguerre Polynomials is different from Bransden & Joachain. My is equivalent to their. (My notation is the more usual one.) 37

38 with Gives (this time I will do it explicitly rather than as an exercise): Now redefine the quantity in summation to make sure every term is the same order: And equate coefficients of (since the equation must hold for any ) 38

39 with So The conventional normalisation is Notice that the polynomials truncate at order if and only if is an integer. Associated Laguerre Polynomials (i.e. k 0) are obtained by differentiation: Clearly these truncate at order if is an integer 39

40 40

41 normalisation For these solutions to be finite everywhere, we need them to truncate must be an +ve integer already taken care of by the angular equation must be an integer must be an integer this also implies Remember our definition 41

42 It is more usual to set (since it is an integer) and write this in terms of the Bohr radius technically this is the reduced Bohr radius - the Bohr radius is We then find that the energy is quantised. The electron can only sit in discrete energy levels with energies: with 42

43 What about the wavefunction? What is the probability distribution for finding an electron a distance r from the nucleus? Recall the probability of finding an electron in volume V is infinitesimal volume element in spherical polars Also recall our normalisation of the spherical harmonics: So the probability density of finding the electron at radius is 43

44 Notice the spectral notation where for 44

45 The hydrogen wavefunctions are, where is a normalization coefficient. To find this coefficient we need You don t need to worry about how to do this integral. It is done using the recurrence relation and an othogonality relation for the Laguerre polynomials. Finally, the electron wavefunction in the hydrogen atom is 45

46 3.4 A summary of the hydrogen quantum numbers and energy levels The electron wavefunction is described by 3 quantum numbers and takes integer values and labels the energy eigenvalues takes integer values and labels the eigenvalues of takes integer values and labels the eigenvalues of The energy levels we predict are (putting in the numbers): ionization energy Notice that we also predict that the energy doesn t depend on. So many electrons can occupy one energy level these groupings of electrons are often called shells and are labelled K, L, M, N, O e.g. the n=0 states are the K-shell etc. 46

47 4. Observations of Hydrogen Spectra 4.1 The Balmer Lines In 1826, Herschel and Fox Talbot noticed that pure samples of hot, glowing material gave off light of very particular colors (i.e. wavelengths) and for each element this set of colours were different. This was observed to also be true for hydrogen, with 4 lines in the visible spectrum: H δ H γ nm nm H β nm H α nm Balmer pointed out that these lines obeyed the formula with and giving four lines. He used his formula to predict another line at nm (k=7) which was then confirmed. 47

48 Rydberg rewrote this as where is the Rydberg constant. We now interpret these lines as photons being emitted by electrons falling from excited states of Hydrogen to the ground state, in this case, the n=3, n=4 and n=5 states falling down to n=2. 48

49 4.3 Other hydrogen emission series Of course, hydrogen has many more such spectral lines the Balmer four were seen first because they are in the visible spectrum. Pfund series Brackett series Paschen series Balmer series Lyman series Each series is named after its discoverer Exercise: Work out the wavelengths for the first (lowest energy) four lines of the Lyman series. 49

50 4.4 The observation of fine splitting In 1887, Michelson and Morley (the same guys who did the aether experiment, and in the same year!) noticed that the H-α line wasn t just one line after all it was split into two nm If we look very closely at the other spectral lines, we will see that they split up too (often into many lines). What have we missed from our calculation? H α nm We need to add in 2 new ingredients to see where this fine-structure comes from: The first relativistic correction to the Schrödinger Equation s Kinetic Energy An interaction between the magnetic field caused by the orbit of the nucleus and the spin of the electron - this is known as the spin-oribit coupling 50

51 4.5 The Stern Gerlach Experiment In 1920, Stern and Gerlach performed an experiment where they passed a beam of silver atoms through a magnetic field. photographic plate slit S N magnets source of Ag They found that the beam slit into two, with half the atoms going up and half going down. The atoms were neutral so this was not caused by the charge (which would anyway have made the atoms move to the side rather than up and down). 51

52 Usually atomic energy levels are filled up with pairs of electrons, one spin-up and one spin-down. This is because the Pauli Exclusion Principle prevents identical fermions occupying the same state, so the electrons need to be different spin to occupy the same level. So every filled level has zero total spin. Energy Silver is special though. It has only one electron in its highest energy shell, so the spin is not zero. We know that magnetic fields can interact with the angular momentum of charged particles. The magnetic field interacts with the spin of the electron in the highest energy level. Half the atoms have a spin-up electron at the top and half have a spin down electron. The interactions for spin-up and spin down have different sign, so the beam is split. This was the first experimental evidence for spin. 52

53 5. Fine Splitting of Hydrogen Energy Levels 5.1 Perturbation Theory The modifications we need to describe fine structure are small modifications to the energy of the system, i.e. they change the Hamiltonian (we will see how later). New Hamiltonian Hamiltonian we had before a small number extra contributions to the Hamiltonian We want to solve just like before, but if we just put in the new terms we will find the Schrodinger Equation in impossible to solve. We get around this using Perturbation Theory if the corrections are small (as we implied by having the factor ) we can assume that the new solutions will be only a small modification of the old solutions and perform a Taylor expansion (in terms of the small quantity) around the old solutions. 53

54 Since the old solutions formed a complete set, we can use them as a basis for our new functions. new solutions satisfying constants to be determined old solutions satisfying If we can determine what are, then we have our solution. But, so (no implicit summation) 54

55 Now multiply by (from the left) and integrate over all space: Remember back on slide 10, we showed that the wavefunctions were orthonormal, so For compactness of notation I will, from now on write this as 55

56 So far, this is exact. Now we will make a series expansion in. We write and. We then find, Equate coefficients of : As long as we have no degenerate eigenvalues (see Dr Campbell s lectures for degenerate perturbation theory), so and 56

57 Equate coefficients of : But now I know, so which tells us that for and 57

58 We can find from the normalisation. If we square this, almost all terms are zero (since and are orthogonal for ) So 58

59 In summary: If then and where and are eigenfunctions and eigenvalues of the unperturbed Hamiltonian. 59

60 5.2 Relativistic corrections to the Kinetic Energy The Schrödinger equation is non-relativistic. The kinetic energy term in the SE is the classical expression, To account for fine structure we must take relativistic corrections into account. Recall that the total energy of a free electron is given by, mass energy The Kinetic energy is 60

61 So, the kinetic energy is classical term 1 st relativistic correction This leads to corrections of the energy levels, To work this out, recall that are solutions of so, and 61

62 We need to be a little bit careful here V depends on r which we can t just pull out of the integral. Writing, we have Some difficult calculus yields: Exercise: Show that 62

63 Putting all this together (with ) gives fine structure constant NB: Not the same as we had earlier 63

64 5.2 Spin-Orbit coupling The derivation of spin-orbit coupling is beyond this course. To do it properly one should use the Dirac Equation which is the relativistic equation for particles with spin. You may meet the Dirac Equation and derive the spin-orbit coupling in Advanced Quantum Mechanics next year. But we can give a heuristic argument. In the rest frame of the electron, the nucleus appears to be accelerating. Remember that accelerating charges produce a magnetic field, so the electron sees a magnetic field of where is the electric field derived from the potential The magnetic field is then Note that V is the potential energy not the potential, so we need to divide by e. 64

65 A magnetic field can interact with the electron s angular momentum - the contribution to the energy is where is the electron s magnetic moment. In the electron s rest frame it has no normal orbital angular momentum, but it does have spin, a form of intrinsic angular momentum. This contributes energy with gyromagnetic ratio is predicted by the Dirac equation (see the Zeeman effect later) spin vector So the naive correction term is: 65

66 This is wrong by a factor of 2! We considered the rest frame of the electron, but forgot to take into account that the frame is also rotating. We could perform a lengthy calculation of this effect, but it is far too long for this course. The result is to modify the correction term to: The extra contribution is known as Thomas Precession. The spin operator is a vector (just like angular momentum) and we can choose electrons wavefunctions which are eigenstates of the operators and with eigenvalues and respectively. We also define a total angular momentum operator and describe the electron wavefunctions with eigenstates of and with eigenvalues and respectively. 66

67 The contribution to the Hamiltonian proportional to doesn t commute with so we can no longer use as a quantum number to describe our states. However, since the new Hamiltonian commutes with and so we can describe our states using and are closely related (superpositions of one another), but we don t have time to examine this relation here. The correction to the energy is: 67

68 5.2 Complete corrections Putting the two corrections together and noting that gives the combined correction from relativity and the spin-orbit coupling: Exercise: Explicitly show the last equality by writing There was a bit of a fudge here (another one!). Notice that is not well defined for either contribution. We actually missed another contribution the Darwin term which is a relativistic correction to the potential. However, it only effects and the above result is correct (coincidentally) for all as long as we insist for 68

69 The energy of a state is now specified by both n and j For example, consider n = 2. Now for we must have, labelled For we have so either labelled and degenerate with the state or labelled Notice the spectral notation is now where for since we also need to quantify. 69

70 Since the transitions are the emission of a spin 1 photon, and we need to conserve angular momentum, not all transitions are allowed. e.g is forbidden because because is forbidden Not to scale! See Dr Campbell s lectures for more on this The Balmer H-α line actually splits into lots of lines (5 lines). At first sight it looks like 2 because the transitions and dominate. 70

71 5.2 The Lamb Shift In the above, we saw that, for example, the and states are degenerate. In 1947, Lamb and Retherford measured these energy levels and found that they differed by. They also observed that the s state had slighly higher energy than predicted. These differences are due to Quantum Electro Dynamics (QED). The exchange of virtual particles in loops causes the electron to be smeared out. This calculation is beyond an undergraduate course, so we will just quote the result: 71

72 6. Hydrogen in an External Magnetic Field In 1896, Zeeman noticed that spectral lines split further if you applied a magnetic field. This should now be obvious to us the magnetic field interacts with the angular momentum, altering the Hamiltonian (total energy). As we saw before (for the discussion of spin-orbit coupling) the appropriate contribution to the energy is of the form: where is the magnetic moment. Since we have to take into account the magnetic moments arising from both the orbital angular momentum and the spin, the appropriate energy contribution is Bohr magneton 72

73 6.1 The Strong Field Case: This requires very strong magnetic fields, such as those in a neutron star. In this case, we can neglect (initially). Choosing the z-axis to be in the direction of the magnetic field, we need to solve, Usual SE Hamiltonian Although this looks hard(!) remember that our solutions to were eigenstates of, so they are also eigenstates of our new equation! Then 73

74 This effect is known as the Zeeman effect. Clearly this will cause the original spectral lines to split into many lines, with the number of new lines depending on the quantum number. or Not to scale! All these lines are separated by the same energy The associated wavelength is and is known as the Larmor frequency. 74

75 6.2 The Weak Field Case:, the anomalous Zeeman Effect If the energy contributed by our field is weak compared to the spin-orbit contribution, we have to use the spin-orbit modified wavefunctions as our basis states. These were, so are not eigenstates of or. We need to use perturbation theory and the correction to the energy is: Let be a direction perpendicular to, i.e., such that. It is easy to see that. 75

76 So and This second contribution is zero. actually precesses around so the average is zero. (This is not obvious!) We now have 76

77 So the correction to the energy levels by the weak field is Landé g-factor This gives further splittings of our spectral lines, e.g. for Not to scale! Basic SE Fine structure Zeeman splitting 77